Problem: Divide the following complex numbers. $ \dfrac{6-3i}{2-i}$
Explanation: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${2+i}$ $ \dfrac{6-3i}{2-i} = \dfrac{6-3i}{2-i} \cdot \dfrac{{2+i}}{{2+i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(6-3i) \cdot (2+i)} {(2-i) \cdot (2+i)} = \dfrac{(6-3i) \cdot (2+i)} {2^2 - (-1i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(6-3i) \cdot (2+i)} {(2)^2 - (-1i)^2} = $ $ \dfrac{(6-3i) \cdot (2+i)} {4 + 1} = $ $ \dfrac{(6-3i) \cdot (2+i)} {5} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({6-3i}) \cdot ({2+i})} {5} = $ $ \dfrac{{6} \cdot {2} + {-3} \cdot {2 i} + {6} \cdot {1 i} + {-3} \cdot {1 i^2}} {5} $ Evaluate each product of two numbers. $ \dfrac{12 - 6i + 6i - 3 i^2} {5} $ Finally, simplify the fraction. $ \dfrac{12 - 6i + 6i + 3} {5} = \dfrac{15 + 0i} {5} = 3 $